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Dr. Antonio R. Quesada, Director Project AMP
Lesson: Systems of Equations
Molly Bordenkircher, Jane Buehler, Stephanie Gonzi, Lester McCurdy
Prior Knowledge: This lesson is based on the idea that students can graph linear equations with and
without technology and are able to solve multi-step equations for a given variable.
Lesson Objective(s):
This lesson is designed to enable students to:
 Solve and interpret the meaning of 2 by 2 systems of linear equations graphically, by
substitution and by elimination, with and without technology. (Patterns, Functions, and Algebra
Standard, Grade 9, Indicator 9)

Solve real-world problems that can be modeled, using systems of linear equations. (Patterns,

Functions, and Algebra Standard, Grade 10, Indicator 11)
Model and solve problems using matrices. (Patterns, Functions, and Algebra Standard, Grade 11,
Indicator 7)

Solve 3 by 3 systems of linear equations by elimination and using technology, and interpret
graphically what the solution means (a point, line, plane, or no solution). (Patterns, Functions, and
Algebra Standard, Grade 11, Indicator 9)

Set up and solve systems of equations using matrices and graphs, with and without technology.
(Patterns, Functions, and Algebra Standard, Grade 12, Indicator 5)
Definition(s):
 A system of equations is a set of equations dealt with simultaneously for which a common
solution, if possible, is sought.
 A solution to system of equations is a point that lies on the graph of each equation in the
system.
Part I: Identifying the solution(s) to a system of equations
1. The equations y = x + 1 and y = ½x + 2 are graphed below to the right. Do the graphs intersect?
If so, name the point of intersection.
2. Verify your intersection point is correct by substituting the coordinates of the intersection point
in for x and y into the original equations. Show your work and explain how your work verifies
that the intersection point is correct
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Dr. Antonio R. Quesada, Director Project AMP
3. The intersection point (2, 3) is the solution to the system of equations in question 1. To solve a
system of equations graphically, graph both equations and see where they intersect. The
intersection point is the solution.
4. Solve Graphically. Graph the following system of equations on your graphing calculator. Paste
a screen shot of your graphs in the space below. (Hint: You will need to solve both equations in
terms of y to graph the equations on your graphing calculator. Show your work for completing
this.)
4x-6y=12
2x+2y=6
y1 = _______________________
y2= _______________________
5. What does the solution to the system of equations from question 4 appear to be? Verify your
intersection point is correct by substituting the coordinates of the intersection point in for x and
y into the original equations. Show your work and explain how your work verifies that the
intersection point is correct.
6. On your graphing calculator, graph the following system of equations and paste a screen shot of
your graphs in the space below.
f1(x) = 2x - 5
f2(x) = -¼x - 1
7. Is the exact solution to the system obvious? Why or why not?
8. When the solution contains non-integer values, the solution can be calculated graphically using
the intersection point(s) tool. Select 6: Points and Lines from the graphing menu and arrow
down to select 3: Intersection Point(s). Select the graph of f1(x) = 2x - 5 by hitting “enter”
when the line is flashing. Next, select the graph of f2(x) = -¼x – 1 by hitting “enter” when the
line is flashing. The solution will appear. Record the solution and paste a screen shot of graphs
with the solution shown below.
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Dr. Antonio R. Quesada, Director Project AMP
9. On a new screen, graph the following system of equations and paste of screen shot of the
graphs below. (Hint: you will need to solve the second equation for y to enter the equation on
your graphing calculator.)
y = 2x + 4
-4x + 2y = 6
10. What relationship exists between the lines? Does the system appear to have a solution? Why or
why not?
The system of equations from question 9 represents one of the three possible cases of systems of
equations, called an inconsistent system.
11. Graph the following systems of equations on your graphing calculator and complete the first
two columns of the chart below. If the system has a solution, give the coordinates of the
solution. (Caution: Lines that appear to be parallel are not always parallel. You may need to
change your viewing window to see a point of intersection.)
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System of
Equations
Dr. Antonio R. Quesada, Director Project AMP
Number of
Solutions
Coordinates of
solution, if one
exists
Case
Name of
System
1.
g(x) = 2x – 0.6
h(x) = -¼x
2.
i(x) = 0.5x + 9
j(x) = ½ x + (18/2)
3.
y=x–1
-x + y = -1
4.
k(x) = 1.5x + 6
l(x) = 3/2x - 4
5.
-5x – y = 4
10x + 2y = 7
6.
m(x) = 4x + 9
n(x) = 3.8x - 8
Three Possible Cases: When we graph a system of two linear equations, one of three things may
happen.
Categorizing Systems by Names – Consistent, Inconsistent, Dependent, and Independent
1. The lines have one point of intersection. The point of intersection is the only solution
of the system. The system is consistent and independent.
2. The lines are parallel. If this is the case, there is no point that satisfies both equations.
The system has no solution. The system is inconsistent.
3. The lines coincide. Therefore, the equations have the same graph and every solution of
one equation is a solution of the other. There is an infinite number of solutions. The
system is consistent and dependent.
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Dr. Antonio R. Quesada, Director Project AMP
This information is summarized in the chart below.
Case
1. lines intersect
2. parallel lines
3. lines coincide
Number of Solutions
One
Zero
Infinitely many
Name of System
Consistent, independent
Inconsistent
Consistent, dependent
Information in question 11 is taken from http://www.jcoffman.com/Algebra2/ch3_1.htm
12. Using the information from above, complete the last two columns of the chart in question 12.
13. Summarize all definitions and concepts concerning systems of linear equations you have
learned thus far.
Part II: Solving Systems of Equations by the Substitution Method
You can solve a system of equations by solving one equation for one of the variables; then substitute
this result into the second equation.
Here is an example of solving a system of equations by substitution.
Find the solution to the following system of equations.
x=y+3
y + 3x = 1
Step 1: The first equation is solved for x. Substitute (y + 3) in for x in the second equation.
y + 3x = 1 becomes y + 3(y + 3) = 1
Step 2: Solve for y.
y + 3y + 9 = 1
4y = -8
y = -2
Step 3: Substitute -2 in for y in the first equation to solve for x.
x = y + 3 becomes
x = -2 + 3
x=1
Step 4: Write the solution as an ordered pair.
The solution to the system of equations is (1, -2).
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Dr. Antonio R. Quesada, Director Project AMP
15. Use substitution to solve the following system of equations:
3x + 4y = -1
6x – 2y = 3
Solve for x in the first equation. Show your work.
Substitute the value you got for x into the second equation, rewrite the equation, and solve for y.
(Hint: Remember to use parentheses.) Show your work.
Now, substitute the value you got for y into either equation and solve for x.
The solution to this system of equations is (x, y) = __________
Check your work by graphing the system of equations on your graphing calculator. Find the
intersection of the two lines. Does it agree with your answer above?
16. Solve each of the following systems.
a. Graph each system of equations.
b. Classify each system as consistent independent, inconsistent, or consistent dependent.
(State the solution if one exists.)
c. Solve the systems by substitution. Be sure to show your work.
d. Notice when solving by substitution in (b) and (c), either an untrue statement results or
an identity results. Compare these answers with the type of system you named in bullet
2.
a. 2x + 3y = 3
12x – 15y = -4
b. x + 3y = 0
2x + 6y = 7
c. 1.4x - .3y = 20
2.8x –. 6y = 40
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Dr. Antonio R. Quesada, Director Project AMP
You can classify a system of linear equations by the number of solutions. Systems that have a unique
solution are independent systems. However, not every system has a unique solution. Systems with an
untrue statement are inconsistent. Systems with an identity are consistent and dependent. Summarize
what types of solution you will get when solving a system by substitution by providing examples of
each type.
17. What will happen if you have a system of 3 equations with 3 variables?
3x – 5y + z = 9
x – 3y – 2z = -8
5x – 6y + 3z = 15
What variable would you solve for in the first equation? _____ What is your
result?_____________________
Now, use this result and substitute it into the second and the third equations. Do you notice that you
now have two equations with two variables? Solve this new system. Substitute the results into an
equation to get the value of the variable that you chose first. The solution to this system of equations is
(x, y, z) = _______________
Verify by substituting the solution into each of the equations.
Part III: Solving Systems of Equations by the Elimination Method
18. Let’s go back to the same system of equations that we used in #15 and find another method to
solve them.
3x + 4y = -1
6x – 2y = 3
Look at the second equation. What value could you multiply the second equation by so that when you
add the first equation to the new second equation, the sum of the y variables would be 0? ___
2( 6x – 2y = 3)______________________________________
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Dr. Antonio R. Quesada, Director Project AMP
Now add the two equations: 3x + 4y = -1
12x – 4y = 6
Solve for x. _____ Substitute this value into either equation and solve for y. Does your answer
correspond to the answer you got by the substitution method, and by graphing?
Could you have first found a value in the first equation to eliminate x? Look again at the first equation.
What value could it be multiplied by in order to eliminate the x variables when the two equations
would be added together?
Here is the work that another student has done to solve a system of equations. Explain what the
student did in each step.
5x + 4 y = 24
3x = 2 + 2y
Step 1. 5x + 4y = 24
3x – 2y = 2
Step 2. 15x + 12y = 72
-15x + 10y = -10
Step 3. 22 y = 62
Step 4. y = 31/11
Step 5. 3x = 2 + 2(31/11)
Step 6. x = 84/33
19. Solve each of the following systems by elimination. Check your answers by graphing, and paste a
screen shot of each graph with the solution shown.
a. 3x – 12y = 25
2x – 4y = 7
b. .5x = .25y
x + 2y = 30
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Dr. Antonio R. Quesada, Director Project AMP
20. Systems in three variables can also be solved by using the elimination method. Work with any
two equations at a time. Look at the following system.
x + 2y + 3z = 5
3x + 2y – 2z = -13
5x + 3y –z = -11
Look at the first two equations. You would probably want to eliminate y first. What value would you
multiply either the first or the second equation by in order to do this?__________
Show your work, and add the two equations.
What variables remain in the sum of the two equations?
Now look at the second and third equations. By what would you multiply the second equation, and by
what would you multiply the third equation, so that you could also eliminate y?
Show your work, and add these two equations.
Now you have two equations with two variables. Continue to use the elimination method and solve for
x and z.
Substitute these two values into an equation and solve for x.
Check your answers by substituting the values for x, y, and z into each equation? What do you notice?
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Dr. Antonio R. Quesada, Director Project AMP
Can you graph these equations and check your results? Why, or why not? (Hint: Remember from
geometry what three non-collinear points determine.) Explain what the intersection of the three
equations might be.
21. Write a system of two equations with two variables in which it would be easier to solve by
substitution. Explain your reasoning. Then solve by both substitution and by graphing. Show your
work and your graph.
22. Write a system of three equations with three variables in which there are infinitely many solutions.
Explain how you know that this will be true in the system you wrote.
23. Explain the significance of solving a system of equations by graphing
Solve using any methods for systems of equations
24.
5x-2y+4z=2
-4x+3y+8z=60
3x+2y-3z=-13
x=_______
y=________
z=________
y=________
z=________
25. 4x+z=24
-7x-6y+25z=0
12x+4y+8z=44
x=_______
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Dr. Antonio R. Quesada, Director Project AMP
26. 6x-4y+16z=9
24z-12x+2y=-1
-8y-4z-6x=0
x=_______
y=________
z=________
Part IV: Solving Systems of Linear Equations with
Row Echelon Form on Matrices
Definition(s):
 A matrix is a rectangular arrangement of numbers
 An augmented matrix is a matrix which contains the coefficients of the variables, the last
column contains the numbers on the right-hand side of the equations.
 Reduced row echelon form is a matrix in row echelon form with every column that has a
leading 1 having 0's in all other positions.
There is no more efficient way to solve a system of linear equations than with Row
Reduction to Echelon Form (rref) on an augmented matrix. There are a variety of
algebraic, geometric, and numerical methods available, but using the rref command
applied to an input augmented matrix handles all possible cases of solving systems of
linear equations. The parts of the following activity will show how to use the TI-Nspire
to deal with a variety of systems, some of which have unique solutions, infinitely many
solutions, or no solutions at all. (NOTE: The TI-83+/84 also have the RREF command
at MATRIX > CALC > RREF to use on an input augmented matrix.)
To get started, look at the example of a system of 3 linear equations containing 3
variables/unknowns.
5 x  6 y  4 z  4
GIVEN the system of 3 linear equations in standard form: 4 x  3 y  5 z  22
7 x  y  6 z  11
The coefficients and constants in the system are going to be entered as a 3 by 4 matrix,
one row for each equation.
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Dr. Antonio R. Quesada, Director Project AMP
On the TI-Nspire CAS, open a document
 1: Add Calculator page
 Choose menu
 8: Matrix & Vector
(NOTE: This is option 7 on the TI-Nspire.)
 4: Reduced Row-Echelon Form to create an rref() command.
 With the cursor inside the parenthesis, choose control, multiplication key, Matrix
 Select 3 rows and tab to enter 4 columns, then tab to OK to create the template to receive the
values for this system
The field of rows and columns are set to receive the coefficients and constants from the system in the
example.
NOTE: It is best to use the tab key to move from one entry field to the next.
The tab key will move through the fields column by column and row by row.
The arrow keys on the NAVPAD are not as automatic as the tab key for navigating the field.
Use the enter key after value entry is complete. The matrix will automatically be row reduced to
echelon form.
The output translates into the following set of equations revealing the solutions to the system of linear
equations for this example:
1x  0 y  0 z  2
0 x  1 y  0 z  3
0 x  0 y  1z  1
or
x2
y  3
z  1
It should be easy to see that the row reduced form gives the solution for the values of x,
y, and z in the last column. Reduction to echelon form indicates that the main
diagonal is populated with 1’s and elements off the diagonal are 0’s when the system
has a unique solution. However, two other possibilities can occur when RREF is
applied to an input augmented matrix representing some system of equations.
The first alternative is a system that has no solutions. A reduced-echelon form that
has a row of zeros in the coefficient section and a non-zero number in the augmentation
column occurs with a system that has no solution. The following is an example of a
system translated into an augmented matrix and the outcome that occurs when RREF is
applied to the system. Follow the steps above for entering this example. Remember to
use the tab key for moving through the entry fields of the matrix you created.
2 x  y  2 z  12
4 x  3 y  3 z  2
6 x  5 y  8z  9
2 1 2  12
should be entered as rref 4 3  3  2  then press enter


6 5  8 9 
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Dr. Antonio R. Quesada, Director Project AMP
9


1 0 2 0 
The output should be 0 1  7 0 . The last row translates into the false statement 0x+0y+0z=1,


0 0 0 1


indicating no solution.
The second alternative is a system with infinitely many solutions. The following is
an example of a system translated into an augmented matrix and the outcome that
occurs when RREF is applied to the system:
x  y  2 z  15
3 x  y  z  20 should be entered as rref
5 x  y  25
1 1  2  15
3 1  1  20 , then press enter


5 1 0  25
1
5

 
1 0 2
2

5
25 
  . The last row translates to the true statement 0x+0y+0z=0
The output should be 0 1 
2
2

0
0
0
0




indicating infinitely many solutions.
In this second case, the variable z is considered a value open to selection or an arbitrary variable.
However, the values of x and y vary along with the choice of value for variable z. Consider the analogy
to the social process of one person becoming engaged to marry another person. With the choice of
future spouse also comes the selection of future mother- and father in-law and a whole bunch of other
in-laws, but the initial selection of a significant other is a more-or-less arbitrary choice. The same is
true with choice of z and the subsequent determination of the values of x and y. To generate members
of the solution set, consider the row-reduced form of the previous example:
1

1
0

2

5
0 1 
2

0 0 0

1
5
5 1
5
x z
x  z
2
2
2 2
2
25 
5
25
25 5
  translates back into the system y  z  
 z
or y  
2
2
2
2 2
0 
z is arbitrary
z is arbitrary


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Dr. Antonio R. Quesada, Director Project AMP
Here are some of the infinitely many solutions that exist for this system based on
random choices of z. All of the solutions are generated by the equations for x and y:
Z
0
1
9
-10
X
5

2
3
2
5
2
Y
25

2
-10
10
25
2
(x,y,z)
5  25
( ,
,0)
2 2
(3,-10,1)
(2,10,9)
5 25
( , ,10)
2 2
EXAMPLES OF OTHER SYSTEMS OF EQUATIONS:
(NOTE: Refer back to the calculator entry instructions on the first page of this activity.)
27. Solve this system of two equations with two unknowns using the method above.
3x  y  5
5 x  3 y  13
28. Arrange the system of two equations with two unknowns into standard form. Find the solution.
5
3
x
2
2
3
5
y x
2
2
y
29. An extra equation is included in the system with two unknowns to be identified. Find the solution.
2 x  y  3
x  2 y  6
7 x  6 y  2
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Dr. Antonio R. Quesada, Director Project AMP
30. Here is a system of 4 equations with 4 unknowns, so your augmented matrix will have 4 rows and
5 columns. Find the solution.
3x  2 y  4 z  w  6
x  4 y  10 z  w  20
 x  y  z  2w  7
6  9 y  z  3w  18
Use the TI-Nspire to check the solution to this system numerically. Provide a screen capture of your
work below.
Courtesy of Paul A. Trogdon, Cary High School, Cary, North Carolina
http://education.ti.com/educationportal/activityexchange/Activity.do?aId=8838&cid=US
Word problems involving systems of equations.
Solve each of the following questions using any of the methods presented in this lesson. Show all work
and state your reasoning for using the method for obtaining the solution.
31. A mechanic earns $15.00 profit per oil change C, and $40.00 profit per brake job j. If the mechanic
wants to earn $220.00 a day and will only do as many oil changes as brake jobs, then what is the
number of oil changes and brake jobs the mechanic has to do to earn $220.00?
32. The Millers just opened a carpet store. Their startup costs were $5000, and their cost of carpet is $8
per square yard. How many square yards of carpet do they need to sell in order to break even if they
sell the carpet for $18 per square yard?
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Dr. Antonio R. Quesada, Director Project AMP
33. You are a human resource director. It is your job to schedule workers for your plant. There are
three workers that are supposed to work a combined 110 hours per week. Worker 1 is supposed to
work 5 hours more than worker 2, and worker 2 is supposed to work 15 hours more than worker 3.
How long should each one work?
34. Beverly has 31 days to complete her quilt for the county fair. The green squares in the quilt can be
sewn at a rate of 5 squares per day, and the red squares at a rate of 8 squares per day. The quilt can
have up to 96 squares total. The green fabric g costs about $1.60 per square and the red fabric r costs
about $2.40 per square. Write the function that describes the cost of the quilt, give the number of days
sewing green squares and red squares, and what the final cost of the quilt?
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